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Pointers are declared using the following sequence.

[storage-class-specifiers] [cv-qualifiers] type-specifiers 
[ms-modifier] declarator ;

where any valid pointer declarator may be used for declarator. The syntax for a simple pointer declarator is as follows:

* [cv-qualifiers] identifier [= expression]

1. The declaration specifiers:

  • An optional storage class specifier. For more information, see Specifiers.

  • An optional const or volatile keyword applying to the type of the object to be pointed to.

  • The type specifier: the name of a type representing the type of the object to be pointed to.

2. The declarator:

  • An optional Microsoft specific modifier. For more information, see Microsoft-Specific Modifiers.

  • The * operator.

  • An optional const or volatile keyword applying to the pointer itself.

  • The identifier.

  • An optional initializer.

The declarator for a pointer to function looks like this:

(* [cv-qualifiers] identifier )( argument-list ) [cv-qualifers]
[exception specification] [= expression];
  • For an array of pointers, the syntax looks like this:

* identifier [ [ constant-expression ] ]
  • However, pointer declarators can be more complex. For more information,see Declarators.

  • Multiple declarators and their initializers may appear together in a single declaration in a comma separated list following the declaration specifier.

A simple example of a pointer declaration is:

char *pch;

The preceding declaration specifies that pch points to an object of type char.

A more complex example is

static unsigned int * const ptr;

The preceding declaration specifies that ptr is a constant pointer to an object of type unsigned int with static storage duration.

The next example shows how multiple pointers are declared and initialized:

static int *p = &i, *q = &j;

In the preceding example, pointers p and q both point to objects of type int and are initialized to the addresses of i and j respectively. The storage class specifier static applies to both pointers.

// pointer.cpp
// compile with: /EHsc
#include <iostream>
int main() {
   int i = 1, j = 2; // local variables on the stack
   int *p;

   // a pointer may be assigned to "point to" the value of
   // another variable using the & (address of) operator
   p = & j; 

   // since j was on the stack, this address will be somewhere
   // on the stack.  Pointers are printed in hex format using
   // %p and conventionally marked with 0x.  
   printf_s("0x%p\n",  p);

   // The * (indirection operator) can be read as "the value
   // pointed to by".
   // Since p is pointing to j, this should print "2"
   printf_s("0x%p %d\n",  p, *p);

   // changing j will change the result of the indirection
   // operator on p.
   j = 7;
   printf_s("0x%p %d\n",  p, *p );

   // The value of j can also be changed through the pointer
   // by making an assignment to the dereferenced pointer
   *p = 10;
   printf_s("j is %d\n", j); // j is now 10

   // allocate memory on the heap for an integer,
   // initialize to 5
   p = new int(5);

   // print the pointer and the object pointed to
   // the address will be somewhere on the heap
   printf_s("0x%p %d\n",  p, *p);

   // free the memory pointed to by p
   delete p;

   // At this point, dereferencing p with *p would trigger
   // a runtime access violation.

   // Pointer arithmetic may be done with an array declared
   // on the stack or allocated on the heap with new.
   // The increment operator takes into account the size 
   // of the objects pointed to.
   p = new int[5];
   for (i = 0; i < 5; i++, p++) {
      *p = i * 10;
      printf_s("0x%p %d\n", p, *p);

   // A common expression seen is dereferencing in combination
   // with increment or decrement operators, as shown here.
   // The indirection operator * takes precedence over the 
   // increment operator ++. 
   // These are particularly useful in manipulating char arrays.
   char s1[4] = "cat";
   char s2[4] = "dog";
   char* p1 = s1;
   char* p2 = s2;

   // the following is a string copy operation
   while (*p1++ = *p2++);

   // s2 was copied into s1, so now they are both equal to "dog"
   printf_s("%s %s", s1, s2);

Sample Output

0x0012FEC8 2
0x0012FEC8 7
j is 10
0x00320850 5
0x00320850 0
0x00320854 10
0x00320858 20
0x0032085C 30
0x00320860 40
dog dog

Another example illustrates the use of pointers in data structures; in this case, a linked list.

// pointer_linkedlist.cpp
// compile with: /EHsc
#include <iostream>
using namespace std;

struct NewNode {
   NewNode() : node(0){}
   int i;
   NewNode * node;

void WalkList(NewNode * ptr) {
   if (ptr != 0) {
      int i = 1;
      while (ptr->node != 0 ) {
         cout << "node " << i++ << " = " << ptr->i << endl;
         ptr = ptr->node;
      cout << "node " << i++ << " = " << ptr->i << endl;

void AddNode(NewNode ** ptr) {
   NewNode * walker = 0;
   NewNode * MyNewNode = new NewNode;
   cout << "enter a number: " << endl;
   cin >> MyNewNode->i;

   if (*ptr == 0)
      *ptr = MyNewNode;
   else  {
      walker = *ptr;
      while (walker->node != 0)
         walker = walker->node;

      walker->node = MyNewNode;

int main() {
   char ans = ' ';
   NewNode * ptr = 0;
   do {
      cout << "a (add node)  d (display list)  q (quit)" << endl;
      cin >> ans;
      switch (ans) {
      case 'a':
      case 'd':
   } while (ans != 'q');




a (add node)  d (display list)  q (quit)
enter a number: 
a (add node)  d (display list)  q (quit)
node 1 = 45
a (add node)  d (display list)  q (quit)
enter a number: 
a (add node)  d (display list)  q (quit)
node 1 = 45
node 2 = 789
a (add node)  d (display list)  q (quit)